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%%%文档的题目、作者与日期
%%\author{王立庆（2019级数学与应用数学1班）}
%\author{学号 \underline{\hspace{4cm}}\,\,\,\, 姓名 \underline{\hspace{4cm}}  }
%%\title{高等代数第六章：向量空间}
%\title{统计软件考试解答 }
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%\date{2023年4月24日}

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{\Large\bf \H 上海立信会计金融学院期终考试卷 } \hspace{0.3cm} {\Large \underline{ B }卷 解答}

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{\large \bf \H 2023 $\sim$ 2024 学年 第 二 学期 }

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{\large \bf \H \underline{ \emph{2022级跨学科跨专业选修课班级} } 《\underline{ \emph{复变函数} }》 课程代码：\underline{ 162250220 }  }

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\begin{enumerate}

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%\newpage 
\item %1
函数 $w=\frac{1}{z}$ 将 $z$ 平面上的单位圆周 $x^2+y^2=1$ 变成 $w$ 平面上的什么曲线？

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{\color{red}解答：从 $z=\frac{1}{w}$ 可得 
\dotfill (\underline{\hspace{0.2cm} 2分 \hspace{0.2cm}})
$$x+iy = \frac{1}{u+iv} = \frac{u-iv}{u^2+v^2}. $$

因此有
\dotfill (\underline{\hspace{0.2cm} 2分 \hspace{0.2cm}})
$$x=\frac{u}{u^2+v^2},\,\,\, y=\frac{-v}{u^2+v^2}.$$

代入 $x^2+y^2=1$ 得到关于 $u,v$ 的等式为
\dotfill (\underline{\hspace{0.2cm} 2分 \hspace{0.2cm}})
$$\frac{u^2}{(u^2+v^2)^2} + \frac{(-v)^2}{(u^2+v^2)^2} =1.$$

两边同乘以 $(u^2+v^2)^2$, 得到 
\dotfill (\underline{\hspace{0.2cm} 2分 \hspace{0.2cm}})
$$u^2 + v^2 = (u^2+v^2)^2.$$

化简得到仍然是单位圆周 
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$$u^2+v^2 = 1.$$

}


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%\newpage 
\item %2
求 $\sin(5i)$ 和 $\cos(5i)$ 的值，并验证 $\sin^2(5i) + \cos^2(5i) =1$. 

\vspace{0.2cm}

{\color{red}解答：根据正弦函数和余弦函数的定义，可得
\dotfill (\underline{\hspace{0.2cm} 6分 \hspace{0.2cm}})
\begin{eqnarray*}
\sin(5i) &=& \frac{e^{i(5i)} - e^{-i(5i)}}{2i} = \frac{ e^{-5} - e^{5} }{2i} = \frac{e^5-e^{-5}}{2}i, \\  
\cos(5i) &=& \frac{e^{i(5i)} + e^{-i(5i)}}{2} = \frac{ e^{5} + e^{-5} }{2}. 
\end{eqnarray*}

代入计算平方和，可得
\dotfill (\underline{\hspace{0.2cm} 4分 \hspace{0.2cm}})
\begin{eqnarray*}
\sin^2(5i) + \cos^2(5i) = \left(\frac{e^5-e^{-5}}{2}i \right)^2 + \left( \frac{ e^{5} + e^{-5} }{2} \right)^2 = 
\frac{e^{10} + e^{-10}-2}{4}(-1) + \frac{e^{10}+e^{-10}+2}{4} = 1. 
\end{eqnarray*}

}


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%\newpage 
\item %3
考虑初等多值函数 $f(z)=\mathrm{Ln}(z)$. 
计算 $f(5+12i)$ 的所有值。

\vspace{0.2cm}

{\color{red}解答：设 $z=x+yi=5+12i$. 则其模长和幅角分别为
\dotfill (\underline{\hspace{0.2cm} 4分 \hspace{0.2cm}})
\begin{eqnarray*}
|z| &=& \sqrt{x^2+y^2} = \sqrt{5^2+12^2}=13, \\
\mathrm{arg}(z) &=& \arctan\frac{y}{x}=\arctan\frac{12}{5}. 
\end{eqnarray*}

根据对数函数的计算公式 $\mathrm{Ln} z = \ln|z| + i\mathrm{arg}(z) + 2k\pi$, 可得
\dotfill (\underline{\hspace{0.2cm} 6分 \hspace{0.2cm}})
\begin{eqnarray*}
\mathrm{Ln}(5+12i) = \ln 13 + i\arctan\frac{12}{5} + 2k\pi, \,\, k=0,\pm 1, \pm 2, \cdots. 
\end{eqnarray*}

}



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%\newpage 
\item %4
设解析函数 $f(z)=\sqrt[4]{z}$ 定义在 $\mathbb{C}-(-\infty,0]$ 上，并且 $f(1)=-1$, 试求 $f(1-i)$ 的值。

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{\color{red}解答：函数 $f(z)$ 的自变量 $z$ 的定义域为去掉负实轴的复平面，即 
\begin{eqnarray*}
z=r(z)e^{i\theta(z)},\,\, r(z)>0,\,\, -\pi< \theta(z) < \pi, 
\end{eqnarray*}
其中 $r(z)$ 和 $\theta(z)$ 分别表示 $z$ 的模长和幅角。函数 $f(z)$ 的所有单值分支为 
\dotfill (\underline{\hspace{0.2cm} 3分 \hspace{0.2cm}})
\begin{eqnarray*}
f_k(z) = \sqrt[4]{r(z)} e^{i \frac{\theta(z)+2k\pi}{4} },\,\, k=0,1,2,3. 
\end{eqnarray*}

由题设条件 $f(1)=-1$,  因为 $r(1)=1, \theta(1)=0$ 以及 $-1=e^{i\frac{0+4\pi}{4}}$, 所以 $k=2$. 
\dotfill (\underline{\hspace{0.2cm} 3分 \hspace{0.2cm}})

所以 
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\begin{eqnarray*}
f(1-i)=f_2(1-i) = \sqrt[4]{r(1-i)} e^{i \frac{\theta(1-i)+4\pi}{4} } 
= \sqrt[8]{2} e^{i \frac{-\frac{\pi}{4}+4\pi}{4} } =\sqrt[8]{2}e^{i\frac{15\pi}{16}}. 
\end{eqnarray*}

}


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%\newpage 
\item %5
求出 $f(z)=z\cos(z)$ 的原函数，并使用原函数计算积分 
$$\int_{-\pi i}^{\pi i} z\cos (z) dz. $$

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{\color{red}解答：函数 $f(z)=z\cos(z)$ 在复平面上解析，
\dotfill (\underline{\hspace{0.2cm} 3分 \hspace{0.2cm}})

一个原函数为 
$F(z)=z\sin(z)+\cos(z)$. 
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因此
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\begin{eqnarray*}
\int_{-\pi i}^{\pi i} z\sin (z) dz &=& [z\sin(z)+\cos(z)] \mid_{-\pi i}^{\pi i} \\ 
&=& [\pi i \sin(\pi i) + \cos(\pi i)] - [-\pi i \sin(-\pi i) + \cos(-\pi i)] \\ 
&=& 0.
\end{eqnarray*}
}


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%\newpage 
\item %6
设 $C$ 为单位圆周 $|z|=1$. 使用解析函数的导数的柯西积分公式，计算 $$\int_C \frac{dz}{z^2(z^2+4)}.$$

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{\color{red}解答：
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函数 $f(z)=\frac{1}{z^2(z^2+4)} = \frac{1}{z^2(z-2i)(z+2i)}$ 在复平面内有三个奇点 $0,2i,-2i$. 
\dotfill (\underline{\hspace{0.2cm} 2分 \hspace{0.2cm}})

函数 $g(z)=\frac{1}{(z-2i)(z+2i)}$ 在圆盘 $|z|\le 1$ 上解析，
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所以由柯西积分公式可得
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\begin{equation*}
g'(0) = \frac{1}{2\pi i} \int_C \frac{g(z)}{(z-0)^2}dz. 
\end{equation*}

因此所求积分为
\dotfill (\underline{\hspace{0.2cm} 3分 \hspace{0.2cm}})
\begin{equation*}
\int_C \frac{dz}{z^2(z^2+4)} 
= 2\pi i g'(0) = 2\pi i \frac{d}{dz} \left( \frac{1}{z^2+4} \right) \Big{\vert}_{z=0}
= 2\pi i \frac{-2z}{(z^2+4)^2} \Big{\vert}_{z=0}
=0.
\end{equation*}


}


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%\newpage 
\item %7
求幂级数 $\sum\limits_{n=1}^{\infty} \cos(in)z^n$ 的收敛半径。

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{\color{red}解答：
通项系数 $c_n=\cos(in) = \frac{e^{iin}+e^{-iin}}{2} = \frac{e^{-n}+e^n}{2}$, 
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根据达朗贝尔公式， 
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$$
\ell = \lim\limits_{n\to\infty} \left\vert \frac{c_{n+1}}{c_n} \right\vert 
= \lim\limits_{n\to\infty} \left\vert \frac{\cos(i(n+1))}{\cos(in)} \right\vert 
= \lim\limits_{n\to\infty} \left\vert \frac{(e^{-n-1}+e^{n+1})/(2)}{(e^{-n}+e^n)/(2)} \right\vert 
=e.
$$
因此该幂级数的收敛半径为 $R=\frac{1}{\ell} = \frac{1}{e}$. 
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}


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%\newpage 
\item %8
将函数 $f(z)=\int_0^z \frac{\sin z}{z} dz$ 展成 $z$ 的幂级数，并指出展式成立的范围。

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{\color{red}解答：
被积函数的泰勒展开为 
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\begin{subequations}
\begin{align}
\frac{\sin(z)}{z} =  \frac{z-\frac{1}{3!}z^3+\frac{1}{5!}z^5-\frac{1}{7!}z^7+ }{z} \cdots 
=  1-\frac{1}{3!}z^2 + \frac{1}{5!}z^4 -\frac{1}{7!}z^6 +  \cdots, \nonumber 
%\tag{$\heartsuit$}
%\label{taylor-4}
\end{align}
\end{subequations}
该幂级数在整个复平面的每一点都收敛，因此在任意一个有界闭集上一致收敛。
\dotfill (\underline{\hspace{0.2cm} 2分 \hspace{0.2cm}})

逐项积分可得
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\begin{subequations}
\begin{align}
\int_0^z \frac{\sin(z)}{z}dz = \int_0^z\left(1-\frac{1}{3!}z^2 + \frac{1}{5!}z^4 -\frac{1}{7!}z^6 +  \cdots \right) 
= z-\frac{1}{3\cdot 3!}z^3 + \frac{1}{5\cdot 5!}z^5 -\frac{1}{7\cdot 7!}z^7 +  \cdots. \nonumber 
%\tag{$\heartsuit$}
%\label{taylor-4}
\end{align}
\end{subequations}
该展式在整个复平面上成立。
\dotfill (\underline{\hspace{0.2cm} 2分 \hspace{0.2cm}})

}


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%\newpage 
\item %9
判断函数 $f(z)=xy^2+ix^2y$ 的可微性与解析性。

{\color{red}解答：
\begin{enumerate}[label=(\arabic*)]
\item  
函数 $f(z)=u+iv$ 在一点可微的充分必要条件是 $u,v$ 在这点可微，而且柯西-黎曼方程在这点成立。
\dotfill (\underline{\hspace{0.2cm} 2分 \hspace{0.2cm}})

现在二元函数 $u=xy^2$ 和 $v=x^2y$ 在整个平面上都是可微的。
\dotfill (\underline{\hspace{0.2cm} 2分 \hspace{0.2cm}})

%为验证柯西-黎曼方程，先
计算偏导数
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\begin{equation*}
u_x = y^2, u_y = 2xy,  
v_x = 2xy, v_y = x^2. 
\end{equation*}
代入柯西-黎曼方程 $u_x=v_y, u_y=-v_x$, 可得
\begin{equation*}
y^2 = x^2, 
2xy = - 2xy. 
\end{equation*}
因此只有在原点 $(x,y)=(0,0)$ 成立柯西-黎曼方程。
\dotfill (\underline{\hspace{0.2cm} 2分 \hspace{0.2cm}})

\item  
因为函数在一点解析，是指在这点的某个领域内处处可微，现在 $f(z)$ 只在 $z=0$ 可微，在其它地方都不可微，所以这个函数处处不解析。
\dotfill (\underline{\hspace{0.2cm} 2分 \hspace{0.2cm}})

\end{enumerate}


}

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%\newpage 
\item %10
设 $C$ 为圆周 $|z|=1$, 分别使用柯西积分公式和柯西积分定理，计算积分 
$$
\int_C \frac{\sin{z}}{z}dz. 
$$

{\color{red}解答：
\begin{enumerate}[label=(\arabic*)]
\item  
将 $f(z)=\sin(z)$ 和 $z_0=0$ 代入柯西积分公式 
$$
f(z_0) = \frac{1}{2\pi i} \int_C \frac{f(z)}{z-z_0}dz, 
$$
可得所求积分为 
\dotfill (\underline{\hspace{0.2cm} 5分 \hspace{0.2cm}})

$$
\int_C \frac{\sin{z}}{z}dz = 2\pi i f(0) = 2\pi i \sin(0) = 0. 
$$

\item  
函数 $g(z)=\frac{\sin z}{z}$ 在原点 $z=0$ 是可去奇点，因此在整个复平面上是解析函数。
根据柯西积分定理，可得任意闭路径上的积分为零，特别地，有
\dotfill (\underline{\hspace{0.2cm} 5分 \hspace{0.2cm}})

$$
\int_C \frac{\sin{z}}{z}dz = 0. 
$$


\end{enumerate}

}

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